Verified Solution • 2025–26 Session moderate

QJEEM25Phf350: JEE Main 2025 Physics - Class 11 JEEM25Phf350 Solution: A bob of mass is suspended at a point by a light string of…

A bob of mass mm is suspended at a point OO by a light string of length ll and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v0v_0 at the point ' A ', the string becomes slack when, the bob reaches at the point ' DD '. The ratio of the kinetic energy of the bob at the points B and C is _________.

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Circular Motion Question 5 English

Multiple Choice Options

  • A 4
  • B 1
  • C 2
  • D 3

Step-by-Step Verified Solution

amp;12mvA2=12mvB2+mghamp;⇒12 m(5 gℓ)=12mvB2+mgℓ2amp;⇒5mgℓ2−mgℓ2=KEBamp;⇒KEB=2mgℓamp;12mvC2=12mvD2+mgℓ2amp;⇒KEC=12mgℓ+mgℓ2=mgℓamp;⇒KEBKEC=2\begin{aligned}& \frac{1}{2} \mathrm{mv}_{\mathrm{A}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgh} \\& \Rightarrow \frac{1}{2} \mathrm{~m}(5 \mathrm{~g} \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mg} \frac{\ell}{2} \\& \Rightarrow \frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}} \\& \Rightarrow \mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell \\& \frac{1}{2} \mathrm{mv}_{\mathrm{C}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{D}}^2+\mathrm{mg} \frac{\ell}{2} \\& \Rightarrow \mathrm{KE}_{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \\& \Rightarrow \frac{\mathrm{KE}_{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2\end{aligned}

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